ATP & Bioenergetics
I. Introduction
II. ATP
A. What is it?
B. How does it function?
Some example data:
Tissue | ATP | ADP | AMP |
Rat liver | 4.7 | 1.5 | 0.6 |
Rat heart | 13.3 | 2.6 | 0.43 |
Blowfly muscle | 14.0 | 3.0 | 0.25 |
The sum of the concentrations of ATP, ADP and AMP will remain relatively
constant in a cell, but depending on the metabolic rate, the percentage of ATP will be
higher.
C. Energy Charge
The relative amount of high energy forms of ATP (ATP, ADP) can be calculated
using the following formula:
EC = 1/2([ADP] + 2[ATP]/[AMP] + [ADP] + [ATP])
Note that if all adenosine phosphates are ATP, EC = 1.0; if all AMP, EC = 0;
and if all ADP or ATP = AMP, EC = 0.5
D. Free energy and ATP
How does the energy in ATP specifically get utilized to power reactions in
metabolism?
First Law: In any process, the total energy of the systems and the surroundings remains constant; energy is not created or destroyed. Energy may be transformed from one form to another
Second
Law: In any process, the entropy of the system and the surroundings increases.
Entropy is often thought of as disorder or randomness.
Free Energy
D
G =DH - TDSwhere, G = free energy; H = enthalpy (total energy contained in chemical
compound); S = entropy; T = absolute temperature
A + B <=>C + D
DG = DG0
+ RT ln ([C][D]/[A][B])
"The Chemists Free Energy"
"The Biologists Free Energy"
R= gas constant (8.31 X 10-3 kJ/mole 0K)
T= absolute temperature
Calculate the standard free energy for a chemical reaction assuming the
following conditions:
Let the reaction go to equilibrium where no net chemical change is occurring
and the free energy change for the system is at a minimum (DG = 0)
Example reaction:
Glucose-1-phosphate <=> Glucose-6-phosphate
This reaction occurs in glycolysis and is catalyzed by the enzyme
phosphoglucomutase. Assume you start with 0.020 M Glucose-1-phosphate, add the enzyme
(along with other appropriate reaction conditions; 25 0C) and allow the
reaction to proceed to equilibrium. The final equilibrium mixture was found to have:
0.001M Glucose-1-phosphate
0.019M Glucose-6-phosphate
The Keq = Glucose-6-phosphate/ Glucose-1-phosphate = 0.019/0.001 =
19
DG0= - RT ln Keq
= - 8.31 X 10-3 kJ/mole 0K x 298 x ln 19
= -7.29 kJ/mole
Note that the value for DG0 is
negative. What does this mean? What type of reaction is this?
From an energetic standpoint, what does this mean?
Is it possible for these reactions to occur in the cell anyway?
E. The Coupling of Chemical Reactions; ATP
For example:
Glucose-1-phosphate <=> fructose-6-phosphate
This reaction is catalyzed by an isomerase and has a DG0 =
+1.86 kJ/mole; this means it will not proceed forward spontaneously under standard
conditions (1M concentrations, etc.).
The Solution:
fructose-6-phosphate + ATP => fructose-1,6-diphosphate +
ADP
DG0 =
-15.8 kJ/mole
Direct coupling by sharing a common enzyme
glucose + phosphate => glucose-6-phosphate (DG0 =
15.4 kJ/mole)
ATP + H2O => ADP + phosphate (DG0 =
-34 kJ/mole)
The synthesis of glucose-6-phosphate would never occur under cellular
conditions unless the reactions are added together:
glucose + ATP => glucose-6-phosphate + ADP
DG0 =
-34 + 15.4 = -18.6 kJ/mole; reaction proceeds spontaneously in the direction written.